Question

necesito ayuda con estas ecuaciones bicuadradas , resolverlas si se pueden.​

Answer 1

Lo que te voy a mostrar probablemente sorprenda a quien te mire la tarea xd

Solo ten en cuenta que:

$\sqrt{ - 1} = i$

y

$a {}^{2} + b {}^{2} = (a - ib)(a + ib)$

1)

$x {}^{4} - 81 = 0 \\ (x {}^{2} ) - (9) {}^{2} = 0 \\ ({x}^{2} + 9)( {x }^{2} - 9) = 0 \\ (x - 3i)(x + 3i)(x + 3)(x - 3) = 0 \\ x = 3i \\ x = - 3i \\ x = 3 \\ x = - 3$

2)

$x {}^{4} + 40 = 0 \\( x {}^{2} + \sqrt{40} i)( {x}^{2} - \sqrt{40} i) = 0 \\ (x + \sqrt{ \sqrt[4]{40}i } \times i)(x - \sqrt{ \sqrt[4]{40}i } \times i)(x + \sqrt{ \sqrt[4]{40}i })(x - \sqrt{ \sqrt[4]{40}i }) = 0 \\ x = \sqrt{ \sqrt[4]{40}i } \times i \\ x = - \sqrt{ \sqrt[4]{40}i } \times i \\ x = \sqrt{ \sqrt[4]{40}i } \\ x = - \sqrt{ \sqrt[4]{40}i }$

3)

${x}^{4} - {x}^{2} = 0 \\ x {}^{2} (x {}^{2} - 1) = 0 \\ x {}^{2} (x + 1)(x - 1) = 0 \\ x = 1 \\ x = 0 \\ x = - 1$

4)

$3 {x}^{4} - 27 {x}^{2} = 0 \\ {x}^{4} - 9x {}^{2} = 0 \\ x {}^{2} (x {}^{2} - 9) = 0 \\ x {}^{2} (x - 3)(x + 3) = 0 \\ x = 3 \\ x = 0 \\ x = - 3$